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In this article we consider the space of functions, that is, the vector space of all maps
of a set
into a vector space
.
Let
be a field,
a
-vector space and
some set.
Then we can define the set of maps of
to
:
Hint
For this definition we have not yet used that
is a vector space. It is sufficient if
is just any set.
On this set we define an addition and a scalar multiplication:
Definition (Vector space operations on
)
The addition
is defined by
for all
and
.
Similarly, we define the scalar multiplication
by
for all
and
.
Hint
For the definition we only need that
is a vector space.
can actually be an arbitrary set (i.e. without an algebraic structure).
Theorem (
is a vector space)
is a
-vector space.
How to get to the proof? (
is a vector space)
We proceed as in the article proofs for vector spaces.
Proof (
is a vector space)
We establish the eight vector space axioms. In the following, we consider
.
Proof step: Associativity of addition
Let
. Then, we have:
This shows the associativity of addition.
Proof step: Commutativity of addition
Let
. Then, we have:
This shows the commutativity of addition.
Proof step: Neutral element of addition
We now need to show that there is a neutral element
exists.
That is,
should hold for all
.
It is obvious that the zero mapping
has this property.
Let
. Then, we have:
This shows that
has a neutral element with respect to addition.
Proof step: Inverse with respect to addition
Let
with
.
We need to show that there exists a
such that
holds.
Since
is a vector space, there exists an inverse
with respect to "
" with
for every
. We now show that
is the inverse of
. We have that:
Furthermore,
is uniquely determined by the well-definiteness of
and the uniqueness of the inverse in
.
Thus we have shown that for any
there exists a
with
.
Proof step: scalar distributive law
Proof step: Vectorial distributive law
Let
and
.
Thus the vector distributive law is also established.
Proof step: Associativity of multiplication
Proof step: unitarity law
Let
. Then, we have:
Thus we have shown the unitary law.
Hint
Some people include the completeness of addition and of scalar multiplication also within the vector space axioms. In our case, they follow from the fact that
is itself a
-vector space. We considered this in the hint after defining the operations.
It must also hold that
is non-empty. This follows directly from the existence of a neutral element with respect to addition.
The set of differentiable functions
an an
-vector space
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In the previous section we showed that the set of all maps of a set
into a
-vector space
is again a
-vector space. We now consider the special case
,
and
. We already know that
is a
-vector space. Hence, we know so far that the set of maps
is an
-vector space.
We now consider the set of differentiable functions
, which is denoted
(as "differentiable").
Theorem
The set of differentiable functions
forms an
-vector space.
Proof
The set of differentiable functions
is a subset of the set of maps
, i.e.
. To show that
also forms an
-vector space, it suffices to show that
is an
-subspace of
.
To do this, we need to establish the 3 subspace axioms.
Proof step: 
The function
is differentiable. So:
.
Proof step: For all
we have that
.
Let
be differentiable, i.e.
. We have shown in Analysis I that the function
is also differentiable. Consequently we have that
.
Proof step: For all
and for all
we have that
.
Let
and
. We have shown in Analysis I that the map
is also differentiable. Thus we have that
.
Thus we have shown that
is an
-subspace of
.
We have already seen that the set of sequences over
forms a vector space with respect to coordinate-wise operations.
So a sequence
with entries in
can be seen as a function
.
In this sense, the sequence space is a special case of the function space
by setting
and
.
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